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An electron moving with the speed 5xx10^...

An electron moving with the speed `5xx10^(6)` per sec is shot parallel to the electric field of intensity `1xx10^(3)N//C`. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of `e= 9xx10^(-31)Kg` charge `= 1.6xx10^(-19)C)`

A

7 m

B

0.7 mm

C

7 cm

D

0.7 cm

Text Solution

Verified by Experts

The correct Answer is:
C
`S=u^2/(2a) = (5 xx10^(6))^(2)/(2xx(qE)/m)=(25 xx10^(12) xx9 xx10^(-31))/(2xx1.6 xx10^(-19) xx10^3)`
`= 0.07m = 7 cm.`
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